Using transistor as switch with low voltage

Basically, I need to monitor a circuit’s output so I can activate another system. The output voltage of active stage is 140mV.

If I am not wrong (from what I read about transistors) 140mV is not enough to put a BJT transistor into saturation (ON mode).

The end goal is turning on an LED and/or a relay, so I really need transistor in the saturation region.

I don’t want to use a micro-controller’s ADC to measure the voltage for such simple device (cost wise)

How can solve the problem in a simplest manner possible?

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Is the output AC or DC?
– JIm Dearden
Sep 14 ’14 at 19:11

  

 

it is DC 140mV output
– dvdmn
Sep 14 ’14 at 19:19

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1 Answer
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Use a comparator.

Just be sure to get one with an input common-mode range that extends down to the negative rail. TL331, for example, might work for you.

  

 

Thanks, it is good to learn something new. just to be clear, lets say I power the comparator with 5v and connect Input- to ground and Input+ to other circuit. Will the output be 5V when Input+ goes to 140mV?
– dvdmn
Sep 14 ’14 at 19:37

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The output is open-collector. So you have to provide a pull-up resistor. The benefit is the output high level is whatever you connect the pull-up to, and it doesn’t have to be the same as the supply voltage.
– The Photon
Sep 14 ’14 at 19:48

  

 

Note you should set a threshold for Input- that’s slightly above ground (say 100mv), or it’ll be ‘on’ all the time. Also you need to check that your comparator can actually handle inputs that close to the rails.
– pjc50
Sep 14 ’14 at 20:14

  

 

I found TL331’s PSpice library, I will try to run some simulations (Planning to learn spice anyway). Thanks for helping
– dvdmn
Sep 14 ’14 at 20:17

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